Hi smiyc86;

Here is my answer to #1, I hope it is what you require.

As you know the way to sum all the digits of any number is to take that number mod 9.
1999 is congruent to 1 mod 9 so 1999^1999 is also going to be congruent to 1 mod 9.
Therefore D = sum of the digits of 1999^1999 = 1

Also for fun:

B=29656
C=28

bobbym

Last edited by bobbym (2009-04-18 10:34:59)

Hi smiyc86;

   Thanks for responding.

   The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9, The sum of the digits of B = 1999^1999 = 29656 (by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1) and C = Sum of the digits of B = 28. D = The sum of the digits of C =10. I assumed you would reduce this 10 down to 1. This is as far as I follow the problem please provide more guidance.

Last edited by bobbym (2009-04-21 18:06:09)

smiyc86;

   You are absolutely right. I was confusing the digital root with the digital sum.
I got the following from:
http://en.wikipedia.org/wiki/Digit_sum

Digital root of 199 =1                                (199 is congruent to 1 mod 9)
Digital sum of 199 =19 but still not 10       (No easy way to compute. Just sum the digits)

Digital root of 84001 = 4                           (84001 is congruent to 4 mod 9)
Digital sum of 84001 = 13                          (No easy way to compute. Just sum the digits)

Based on what I think you require, I am stuck on following the rules for digital sums so:

A=1999^1999
B=29656
C=28
D=10

Thanks a lot for listening, please respond.

Last edited by bobbym (2009-04-23 20:50:00)

Hi smiyc86;

This process:

A=1999^1999
B=29656
C=28
D=10
E=?

is the definition of the digital root. That is computed by 1999 is congruent to 1 mod 9 so 1^1999 =1. This saves all the computing. E=1. How was that?

I was afraid that you would ask that. I did the calculation using a computer and added every digit up. This is as far as I know, and that wikipedia site agrees, the only way to get the digital sum. The digital root can be done by congruences.

Last edited by bobbym (2009-04-22 18:57:51)

Hi smiyc86;

Found this page:
http://nz.answers.yahoo.com/question/index?qid=20080620225029AAeW6Uh

with this proof:

We know that Iterative Sum of Digits = n mod 9( also called the digital root)

D = (1999^1999) mod 9 = (1999 mod 9)^1999

We know:

1) 1999 mod 9 = 1
2) 1^f = 1, for any value of f.

D = 1^1999 = 1

This is what I have been saying all along.

check the solution of manjyome... her only mistake is she found out D as the answer whereas the reqd answer is the sum of digits of D.

one more thing

the number of digits in 'n' is (characteristic of log n) + 1

i have already given a proof to it